A force F=t is applied to block A as shown in | vedclass

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(C) $1$. Initially,for block $A$,the applied force $F=t$ is balanced by static friction $f_{rA}$. The block $A$ remains at rest until $F = \mu_s mg$,i

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(C) $1$. Initially,for block $A$,the applied force $F=t$ is balanced by static friction $f_{rA}$. The block $A$ remains at rest until $F = \mu_s mg$,i.e.,$t = \mu_s mg$. During this time,the tension $T$ in the string is $0$,so block $B$ remains at rest and the friction force on $B$ is $0$.$2$. When $t > \mu_s mg$,block $A$ starts to move. The tension $T$ in the string increases as $T = F - f_{rA} = t - \mu_k mg$ (assuming kinetic friction $\mu_k mg$ acts on $A$).$3$. Block $B$ will start to move when the tension $T$ exceeds the maximum static friction force on $B$,which is $\mu_s mg$. So,$T = t - \mu_k mg = \mu_s mg$,which gives $t = \mu_s mg + \mu_k mg$.$4$. For $t  \mu_s mg$,the friction on $B$ is $0$ until $t = \mu_s mg$,then it increases linearly with $t$.$5$. Once $t$ reaches the threshold where $B$ moves,the friction force on $B$ drops from its maximum static value $\mu_s mg$ to the kinetic friction value $\mu_k mg$ and remains constant.$6$. Comparing this behavior with the given options,graph $C$ correctly represents this: zero friction until $t = \mu_s mg$,then increasing friction,a peak at the threshold,and finally a constant kinetic friction value.

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